Definition:Cantor Set

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As a Limit of Intersections

Define, for $n \in \N$, subsequently:

$k \left({n}\right) := \dfrac {3^n - 1} 2$
$\displaystyle A_n := \bigcup_{i \mathop = 1}^{k \left({n}\right)} \left({\frac {2 i - 1} {3^n} \,.\,.\, \frac {2 i} {3^n} }\right)$

Since $3^n$ is always odd, $k \left({n}\right)$ is always an integer, and hence the union will always be perfectly defined.

Consider the closed interval $\left[{0 \,.\,.\, 1}\right] \subset \R$.


$\mathcal C_n := \left[{0 \,.\,.\, 1}\right] \setminus A_n$

The Cantor set $\mathcal C$ is defined as:

$\displaystyle \mathcal C = \bigcap_{n \mathop = 1}^\infty \ \mathcal C_n$

From Ternary Representation

Consider the closed interval $\left[{0 \,.\,.\, 1}\right] \subset \R$.

The Cantor set $\mathcal C$ consists of all the points in $\left[{0 \,.\,.\, 1}\right]$ which can be expressed in base $3$ without using the digit $1$.

From Representation of Ternary Expansions, if any number has two different ternary representations, for example:

$\dfrac 1 3 = 0.10000 \ldots = 0.02222$

then at most one of these can be written without any $1$'s in it.

Therefore this representation of points of $\mathcal C$ is unique.

As a Limit of a Decreasing Sequence

Let $I_c \left({\R}\right)$ denote the set of all closed real intervals.

Define the mapping $t_1: I_c \left({\R}\right) \to I_c \left({\R}\right)$ by:

$t_1 \left({\left[{a \,.\,.\, b}\right]}\right) := \left[{a \,.\,. \frac 1 3 \left({a + b}\right)}\right]$

and similarly $t_3: I_c \left({\R}\right) \to I_c \left({\R}\right)$ by:

$t_3 \left({\left[{a \,.\,.\, b}\right]}\right) := \left[{\frac 2 3 \left({a + b}\right) \,.\,. b}\right]$

Note in particular how:

$t_1 \left({\left[{a \,.\,.\, b}\right]}\right) \subseteq \left[{a \,.\,.\, b}\right]$
$t_3 \left({\left[{a \,.\,.\, b}\right]}\right) \subseteq \left[{a \,.\,.\, b}\right]$

Subsequently, define inductively:

$S_0 := \left\{{\left[{0 \,.\,.\, 1}\right]}\right\}$
$S_{n+1} := t_1 \left({C_n}\right) \cup t_3 \left({C_n}\right)$

and put, for all $n \in \N$:

$C_n := \displaystyle \bigcup S_n$

Note that $C_{n+1} \subseteq C_n$ for all $n \in \N$, so that this forms a decreasing sequence of sets.

Then the Cantor set $\mathcal C$ is defined as its limit, i.e.:

$\mathcal C := \displaystyle \bigcap_{n \mathop \in \N} C_n$

These definitions are all equivalent, as shown on Equivalence of Definitions of Cantor Set.

Also known as

Some sources refer to the Cantor set as Cantor's discontinuum.

Also see

  • Results about the Cantor set can be found here.

Source of Name

This entry was named for Georg Cantor.